Lecture 4: Normalized Functions of Bounded Variation; Regular Borel Measures; the Riesz–Markov–Kakutani Theorem;

Home , Borel measure, Bounded variation

Functional analysis Lecture 4: Normalized functions of bounded variation; regular Borel measures; the Riesz–Markov–Kakutani theorem;

Remark 3.10. Assume that the functional Λ above is positive (i.e. Λf ≥ 0 for f ≥ 0). We observe that the extension Λe is also positive and hence the constructed function f is then monotone increasing. Indeed, suppose that there is g ∈ `∞[0, 1] with g(t) ≥ 0 for each t ∈ [0, 1] but Λeg < 0. By normalizing, we can assume that kgk∞ = 1, thus 0 ≤ g(t) ≤ 1 for each t ∈ [0, 1]. Let 1 be the constant 1 function. As Λ is positive and linear, it is monotone (i.e. Λg ≤ Λh if g ≤ h), and hence Λ1 = kΛk. We have k1 − gk∞ ≤ 1 and

Λ(e 1 − g) = Λ1 − Λeg = kΛk − Λeg > kΛk = kΛek, which gives a contradicition. Therefore, positive continuous functionals on C[a, b] correspond to increasing functions on [a, b], while general continuous functionals correspond to functions of bounded variation.

Remark 3.11. Since every f ∈ BV([a, b]) is the diﬀerence of two increasing functions, we see that every Λ ∈ (C[a, b])∗ can be written as

Λ = Λ+ − Λ−, where Λ+, Λ− are positive. (3.1)

The conclusion of Remark 3.11 can be justified without appealing to Riesz’ Theo- rem 3.9. To this end, let us use a more general language which will be also useful in the sequel. By an ordered vector space we mean any vector space E over R equipped with a partial order relation ≥ compatible with the algebraic operations in the sense that x ≥ y implies x + z ≥ y + z for all x, y, z ∈ E and x ≥ y implies λx ≥ λy for all x, y ∈ E, λ ≥ 0. We denote by E+ the positive cone of E defined by {x ∈ E : x ≥ 0}. An ordered vector space which is also a lattice with respect to the given order is called a Riesz space or a vector lattice. If E is a Riesz space and ∨, ∧ denote the lattice operations, then for any vector x ∈ E we define its positive part x+, negative part x− and its modulus |x| by

x+ = x ∨ 0, x− = (−x) ∨ 0, |x| = x ∨ (−x).

We then have x = x+ − x− and |x| = x+ + x−, so this gives a decomposition of x as the diﬀerence between two positive elements. A Riesz space is called Archimedean if whenever 0 ≤ nx ≤ y for all n ∈ N and some y ∈ E+, then x = 0. If E is a Riesz space and a normed space with a norm k·k satisfying the condition

|x| ≤ |y| =⇒ kxk ≤ kyk for all x, y ∈ E, then E is called a normed Riesz space, and in the case where the norm is complete, we call E a Banach lattice. There are plenty of examples of Banach lattices: Rn with the pointwise ordering and any of the norms k·kp (1 ≤ p ≤ ∞), the space C(X) of continuous

1 functions, and Cb(X) of bounded continuous functions on any topological space X, also considered with the pointwise ordering and the supremum norm, as well as Lp(µ)-spaces under the a.e. pointwise ordering. If E is a normed Riesz space, then its dual E∗ can be ordered in a natural way by deﬁning f ≥ g if and only if f(x) ≥ g(x) for every x ∈ E+ (f, g ∈ E∗). We will see (Lemma 3.18 below) that in this way we equip E∗ with a Banach lattice structure1. In particular, the decomposition (3.1) is possible in the Banach lattice (C[a, b])∗. Actually, decomposition (3.1) holds true with a further requirement that

kΛk = kΛ+k + kΛ−k and in a much more general setting of the so-called order unit normed Riesz spaces (see Problem 3.22). For C[a, b] we will obtain that equality with the aid of Hahn’s decomposition theorem. As we know that every continuous linear functional on the real space C[a, b] is rep- resented by a function f ∈ BV[a, b], let us consider the problem of uniqueness of this representation. Obviously, f is not unique as stated in Theorem 3.9, because adding any constant function to f does not change the Riemann–Stieltjes integral. So, one normalization would be to require that e.g. f(a) = 0. By Jordan’s decomposition, we have f = g −h for some increasing functions g and h, and therefore f has at most countably many points of discontinuity in (a, b). Observe that by modifying the values of f at each of these points, but keeping the variation bounded, we do not change the value of the integral in formula (3.7). Indeed, any g ∈ C[a, b] is Riemann–Stieltjes integrable, hence its integral can be approximated by any sequence of integral sums over partitions with diameters converging to zero2. In particular, we can avoid each point of discontinuity of f (except the endpoints) in these partitions. Consequently, another normalization condition which does not aﬀect the validity of (3.7) is that f(t+) = f(t) for each t ∈ (a, b), i.e. f is right- continuous inside the interval (a, b) (the choice of the left or right limit is arbitrary). Every f ∈ BV([a, b]) satisfying the above two normalization conditions is called a normalized function of bounded variation, and we denote the class of all such functions by

• NBV([a, b]) = {f ∈ BV([a, b]): f(a) = 0 and f is right-continuous in (a, b)}. By modifying the representing function f in Theorem 3.9 as described above, we can guarantee that it belongs to the class NBV([a, b]). Moreover, it turns out that in this class the function f is unique. We will obtain this as a particular case of the ‘uniqueness part’ of Theorem 3.16 below. Before proceeding to that result, let us build a ‘bridge’ between the special Riesz Theorem 3.9 and the much more general Riesz–Markov–Kakutani Theorem 3.16. Given any function f ∈ BV([a, b]), we define a measure µf on the field F consisting of finite unions of the intervals |y, x](a ≤ y < x ≤ b) which are all right-closed and left-open unless y = a in which case we take the closed interval [a, x]. Namely:

• µf ([a, x]) = f(x) − f(a),

• µf ((y, x]) = f(x) − f(y) for a < y < x ≤ b.

1An interested reader may consult e.g. [C.D. Aliprantis, K.C. Border, Inﬁnite dimensional analysis. A hitchhiker’s guide, 3rd edition, Springer 2006; Ch. 8 and 9]. 2See [W Rudin, Principles of mathematical analysis; Thms. 6.6 and 6.8].

2 Notice that the measure µf is ﬁnitely additive and satisﬁes

n n X [ µf (E) = (f(bj) − f(aj)) for E = |a1, b1] ∪ (aj, bj] ∈ F, j=1 j=2 where a ≤ a1 < b1 ≤ a2 < b2 ≤ ... ≤ an < bn ≤ b. Moreover, µf is bounded because f is of bounded variation. But, as we will see, µf has also another nice property of being regular, provided that f belongs to the class NBV([a, b]).

Definition 3.12. Let F be a field of subsets of a topological space X and µ: F → K be a finitely additive set function, where K ∈ {R, C}. We say µ is regular if for every E ∈ F and ε > 0 there exist a set F ∈ F with F ⊆ E and a set G ∈ F with int G ⊇ E satisfying

|µ(C)| < ε for every C ∈ F,C ⊆ G \ F.

If f ∈ NBV([a, b]), it is possible to extend µf to a regular σ-additive measure defined on all Borel subsets of [a, b]. It follows from the following general result on extensions of measures. (Actually, we only need a simplified version of this result, as the measure µf is trivially σ-additive in the sense that there are no countably infinite, but not finite, unions of the intervals |c, d] in F.) Extension theorem. Let F be a field of subsets of some topological space X and let µ: F → K be a bounded, regular, finitely additive set function, where K ∈ {R, C}. Then, µ has a unique regular, σ-additive extension µe: σ(Σ) → K to the σ-algebra σ(F) generated by F. For the proof see e.g. [N. Dunford, J.T. Schwartz, Linear operators (vol. 1: General theory), Wiley–Interscience 1988; Ch. III.5]. This theorem is a combination of two classical measure theory results:

• the Alexandroff theorem which says that every bounded regular finitely additive (real- or complex-valued) set function on a field of subsets of a compact topological space is σ-additive;

• the Carath´eodory extension theorem which guarantees that positive σ-additive measures deﬁned on a ring R can be extended to positive σ-additive measures on the σ-algebra generated by R. (It should also be noted that a measure is regular if and only the positive and negative parts of its real and imaginary parts are regular.) Lemma 3.13. If f ∈ BV([a, b]), then for every t ∈ [a, b) we have

lim V (f, [t, t + ε]) = 0. ε→0+ Proof. (classes) Now, assume that f ∈ NBV([a, b]). Since f is right-continuous on (a, b), we have limε→0 f(t + |ε|) = f(t) for every t ∈ (a, b) and it follows that for every interval |c, d] ∈ F,

n n X Sn o V (f, [c, d]) = sup |µf (Ij)|: Ij ∈ F,Ij ∩ Ik = ∅ for j 6= k, j=1Ij ⊆ |c, d] . j=1

3 (The right-hand side deﬁnes the variation of µf on the set |c, d]; we will introduce this notion later on.) Plainly, we can extend this formula to any set E ∈ F in the sense that if E = I1 ∪ ... ∪ Im, with Ij = |cj, dj] ∈ F (1 ≤ j ≤ m), then the left-hand side is Pm j=1 V (f, [aj, bj]), whereas the right-hand side is the variation of µf on E, denoted by |µf |(E). For any such E we can assume that a ≤ c1 < d1 < c2 < d2 < . . . < cm < dm ≤ b. Pick 0 < ε < min (dj − cj), min (cj+1 − dj) 1≤j≤m 1≤j

(The interval (c1, d1 + ε] should be replaced by [a, d1 + ε] in the case where c1 = a.) Then, in view of the formula above and Lemma 3.13, we have

m X |µf |(E \ E1(ε)) = V (f, [cj, cj + ε]) −−−−→ 0 ε→0+ j=1 and m X |µf |(E2(ε) \ E) = V (f, [dj, dj + ε]) −−−−→ 0. ε→0+ j=1

This shows that µf is regular and thus the extension theorem applies. For σ-additive measures we shall use the following common definition of regularity. It does not completely coincide with Definition 3.12 in general, but it does if X is a locally compact, σ-compact3 Hausdorff space. In this case, a positive Borel measure on X is regular if and only if for every Borel set E ⊆ X and each ε > 0 there exist a closed set F and an open set V such that F ⊆ E ⊆ V and µ(V \ F ) < ε (classes).

Deﬁnition 3.14. Let µ be a positive Borel measure on a locally compact Hausdorﬀ space X. A Borel set E ⊆ X is called outer regular (resp. inner regular) if

µ(E) = inf µ(V ): E ⊆ V,V is open

(resp. µ(E) = sup µ(K): K ⊆ E,K is compact ). The measure µ is called regular if every Borel subset of X is both outer and inner regular.

We have proved that any f ∈ NBV([a, b]) gives rise to a bounded, regular, ﬁnitely additive measure on F to which we can apply the extension theorem. Also, it is not diﬃcult to see that, in the converse direction, the regularity of µf implies that f is right- continuous on (a, b). Summarizing, we have the following result.

Proposition 3.15. Let f ∈ BV([a, b]) (real- or complex-valued), f(a) = 0, let F be the field of finite unions of the intervals of the form [a, d] and (c, d] (a < c < d ≤ b), and let µf : F → R be the finitely additive (real- or complex-valued) measure associated with f as above. Then, µf is regular if and only if f ∈ NBV([a, b]) in which case µf can be uniquely extended to a Borel, regular, σ-additive measure.

3X is σ-compact if it is a countable union of compact sets.

4 In the Riesz–Markov–Kakutani theorem we deal with positive functionals on the space of compactly supported continuous functions. We use the following notation:

• supp(f) = {x ∈ X : f(x) 6= 0}

• Cc(X) = {f : X → K | supp(f) is compact} Also, when given a topological space X, we write:

• K ≺ f if K ⊆ X is compact, f ∈ Cc(X), 0 ≤ f ≤ 1 and f|K = 1,

• f ≺ V if V ⊆ X is open, f ∈ Cc(X), 0 ≤ f ≤ 1 and supp(f) ⊂ V ,

• K ≺ f ≺ V if K ≺ f and f ≺ V . Two basic topological tools which are essential in the proof are the following results: Urysohn’s lemma. Let X be a locally compact Hausdorﬀ space, V ⊆ X be an open set and K ⊂ V a compact set. Then, there exists a function f ∈ Cc(X) such that K ≺ f ≺ V.

Partition of unity. Let X be a locally compact Hausdorﬀ space. Suppose V1,...,Vn ⊆ X are open set and K is a compact set, K ⊂ V1 ∪...∪Vn. Then, there exist functions hi ≺ Vi (for 1 ≤ i ≤ n) satisfying

h1(x) + ... + hn(x) = 1 for every x ∈ K.

In what follows, the space Cc(X) can consist of real- or complex-valued continuous functions with compact support. Correspondingly, Λ can be real-valued and R-linear, as well as complex-valued and C-linear. However, the crucial assumption is that Λ is positive. Notice that the measure µ produced by the theorem below can be inﬁnite. For k R +∞ example, one can think of X = R (or, more generally R ) and Λf = −∞ f(x) dx in which case the measure µ is the Lebesgue measure on R. Theorem 3.16 (Riesz–Markov–Kakutani). Let X be a locally compact Hausdorﬀ space and Λ be a positive linear functional on Cc(X). Then, there exist: a σ-algebra M in X which contains all Borel subsets of X, and a positive measure µ on M satisfying the following conditions: Z (a)Λ f = f dµ for every f ∈ Cc(X); X (b) for every compact set K ⊂ X we have µ(K) < ∞; (c) for every E ∈ M, µ(E) = inf µ(V ): E ⊆ V,V is open ;

(d) for every open set E and every E ∈ M with µ(E) < ∞,

µ(E) = sup µ(K): K ⊆ E,E is compact ;

(e)( X, M, µ) is complete, i.e. if E ∈ M, µ(E) = 0 and A ⊂ E, then A ∈ M. Moreover, the measure µ is unique in the class of positive measures on M satisfying conditions (a)–(d).

5 Proof. The proof consists of three main steps. First, we show that there is at most one measure µ satisfying the announced properties. Next, we provide definitions of µ and M, together with an auxiliary class MF . We do it in such a way that assertion (c) follows from the very definition, while (e) is trivial. Finally, and this will be the toughest part, we shall show that M is a σ-algebra containing all Borel set, µ is σ-additive on M and satisfies (a), (b) and (d). For clarity, the last, most complicated step will be split into ten parts. Uniqueness. If M and µ satisfy (c) and (d), then µ is completely determined by its values on compact sets. Hence, if ν is another positive measure on M satisfying (a)–(d), it is enough to show that µ(K) = ν(K) for every compact set K ⊂ X. For any ε > 0, conditions (b) and (c) imply that there exists an open set V such that K ⊂ V and ν(V ) < ν(K)+ε. By Urysohn’s lemma, there is a function f with K ≺ f ≺ V , and hence Z Z Z Z µ(K) = 1K dµ ≤ f dµ = Λf = f dν ≤ 1V dν = ν(V ) < ν(K) + ε. X X X X Therefore, µ(K) ≤ ν(K) and, by symmetry, µ(K) = ν(K). Construction of M and µ. We define:

µ(V ) = sup Λf : f ≺ V for any open set V ⊆ X, (3.2)

µ(E) = inf µ(V ): E ⊆ V,V is open for an arbitrary E ⊆ X. (3.3) Notice these two definitions are compatible. For, if E is open, then for any open set U ⊇ E, the values of µ defined by (3.2) satisfy µ(E) ≤ µ(U). Hence, µ(E) defined by formula (3.3) is the same as that defined by (3.2). Next, define MF to be the collection of all sets E ⊂ X such that

µ(E) < ∞ and µ(E) = sup µ(K): K ⊆ E,K is compact .

Finally, let M = E ⊂ X : E ∩ K ∈ MF for every compact set K ⊂ X . Observe that by formula (3.2) we have guaranteed condition (c). Of course, the function µ is monotone, i.e. A ⊆ B implies µ(A) ≤ µ(B) and it implies that condition (e) is also satisﬁed. Proof of properties (a), (b) and (d).

∞ Part 1. µ is an outer measure, that is, for any sequence (Ei)i=1 of subsets of X we have ∞ ∞ [ X µ Ei ≤ µ(Ei). i=1 i=1

First, we show this inequality for two open sets. So, assume V1,V2 ⊆ X are open and pick a function g with g ≺ V1 ∪ V2. Using the partition of unity, we produce functions h1, h2 such that h1 ≺ V1, h2 ≺ V2 and h1(x) + h2(x) = 1 for each x ∈ supp(g). Hence, h1g ≺ V1, h2g ≺ V2 and g = h1g + h2g which yields

Λg = Λ(h1g) + Λ(h2g) ≤ µ(V1) + µ(V2).

6 Passing to supremum over all the functions g with g ≺ V1 ∪ V2, we obtain µ(V1 ∪ V2) ≤ µ(V1) + µ(V2). The rest of the proof is left for the classes.

Part 2. For any compact set K ⊂ X we have K ∈ MF and µ(K) = inf Λf : K ≺ f .

Consider any function f with K ≺ f and α ∈ (0, 1). Define Vα = {x ∈ X : f(x) > α} and note that K ⊂ Vα and αg ≤ f provided that g ≺ Vα. Hence, −1 µ(K) ≤ µ(Vα) = sup Λg : g ≺ Vα ≤ α Λf. Passing to the limit as α → 1, we get µ(K) ≤ Λf < ∞. Therefore, K satisfies the requirements of the definition of MF , thus K ∈ MF . We have established assertion (b). Part 3. Every open set V ⊆ X satisfies µ(V ) = sup µ(K): K ⊆ V,K is compact , and hence MF contains all open sets V for which µ(V ) < ∞. (classes)

∞ Part 4. Let (Ei)i=1 be a sequence of mutually disjoint sets from the class MF . Then,

∞ ∞ [ X µ Ei = µ(Ei). i=1 i=1 S∞ S∞ Moreover, if µ( i=1Ei) < ∞, then i=1Ei ∈ MF . We will prove that

µ(K1 ∪ K2) = µ(K1) + µ(K2) for compact K1,K2 ⊂ X,K1 ∩ K2 = ∅. (3.4)

Fix any ε > 0. By Urysohn’s lemma, there is a function f ∈ Cc(X) such that f|K1 = 1 and f|K2 = 0. In view of Part 2, we can pick a function g with

K1 ∪ K2 ≺ g and Λg < µ(K1 ∪ K2) + ε.

Notice that K1 ≺ fg and K2 ≺ (1 − f)g. Since in Part 2 we have proved that µ(K) ≤ Λh for any compact K and any h with K ≺ h, we have

µ(K1) + µ(K2) ≤ Λ(fg) + Λ(g − fg) = Λg < µ(K1 ∪ K2) + ε. Therefore, we obtain formula (3.4) by the fact that ε > 0 was arbitrary. The rest of the proof is left for the classes.

Part 5. For every E ∈ MF and any ε > 0 there exist a compact set K and an open set V such that K ⊆ E ⊆ V and µ(V \ K) < ε. (classes)

Part 6. If A, B ∈ MF , then A ∪ B,A ∩ B,A \ B ∈ MF . (classes) TBC